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pav-90 [236]
3 years ago
10

If you were to separate all of the electrons and protons in 1.00 g (0.001 kg) of matter, you’d have about 96,000 C of positive c

harge and the same amount of negative charge. If you placed these charges 1.00 m apart, how strong would the attractive forces between them be?
Physics
1 answer:
Nookie1986 [14]3 years ago
8 0

Answer:

force = 82.94 × 10^{18}  N

Explanation:

given data

positive charge q1 = 96,000 C

negative charge q2 = -96,000 C

distance r = 1 m

solution

we know here that force between two charge that is express as

force = \frac{K \ |q1\times q2|}{r^2}    ......................1

put here value and we will get forces between them  

force = \frac{9\times 10^9 \times  \ |96000 \times (-96000)|}{1^2}  

force = 82.94 × 10^{18}  

so forces between them 82.94 × 10^{18}  N

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